The simplified circuit of the circuit given in question as follows: 
The equivalent capacitance between C & D capacitors of 2μF,5μF and 5μF are in parallel. ∴CCD=2+5+5=12μF(∵ In parallel grouping Ceq=C1+C2+….+Cn ) Similarly equivalent capacitance between E &BCEB =4+2=6μF Now equivalent capacitance between A & B Ceq1=61+121+61=125 ⇒Ceq=512=2.4μF(∵ In series grouping, Ceq1=C11+C21+………+Cn1 )